If AA + BB + CC = ABC, then what is the value of A+B+C = ?

Concept: A two-digit number with both digits the same, such as AA, equals 11 x A, because AA = 10A + A = 11A. So the sum of three such repdigit numbers is AA + BB + CC = 11(A + B + C).

1. Write AA + BB + CC as 11(A + B + C), since AA = 11A, BB = 11B, and CC = 11C.

2. Write the target as ABC = 100A + 10B + C (its usual place-value expansion).

3. Equate the two expressions: 100A + 10B + C = 11A + 11B + 11C, which simplifies to 89A = B + 10C.

4. Since B and C are single digits (0-9), the largest possible value of B + 10C is 9 + 90 = 99. So 89A is at most 99, which forces A = 1 (A = 2 would need 89 x 2 = 178, which is impossible).

5. Substitute A = 1: B + 10C = 89. For B to stay a single digit (0-9), 10C must be 80, so C = 8 and B = 9.

6. So A = 1, B = 9, C = 8.

Cross-check: Substituting back, AA + BB + CC = 11 + 99 + 88 = 198, and ABC with A=1, B=9, C=8 is indeed 198 -- the equation holds.

So A + B + C = 1 + 9 + 8 = 18.